Consider a system of linear equations A x = b where A is an n × n matrix. Suppose that b is a non-zero vector such that A t b = 0. Which is true about any such system?I am given 5 choices for what this means: The system has infinitely many solutions, is inconsistent, is consistent, is over determined, or is in row echelon form.Unfortunately, I have yet to completely rule out a single one. I don't think it can be over determined, because then wouldn't the equations be undefined based on the fact that if there were more equations than unknowns then A x would not be possible. I don't recall any theorems discussing whether the system is consistent or inconsistent using the transposition and b, which stems to the infinitely many solutions.

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Consider a system of linear equations   A  x  =  b where   A is an   n  ×  n matrix. Suppose that b is a non-zero vector such that       A    t    b  =  0. Which is true about any such system?I am given 5 choices for what this means: The system has infinitely many solutions, is inconsistent, is consistent, is over determined, or is in row echelon form.Unfortunately, I have yet to completely rule out a single one. I don&#039;t think it can be over determined, because then wouldn&#039;t the equations be undefined based on the fact that if there were more equations than unknowns then   A  x would not be possible. I don&#039;t recall any theorems discussing whether the system is consistent or inconsistent using the transposition and b, which stems to the infinitely many solutions.

Odin Jacobson · Accepted Answer

Suppose   A  x  =  b has a solution       x    0  . Then       A    t    b  =      A    t    A      x    0    =  0. This implies either       x    0    t        A    t    A      x    0    =      |        |    A      x    0        |              |        2    =  0. So   b needs to be the zero vector.

polivijuye · Accepted Answer

I&#039;ll assume   A has real entries. A useful fact in this case is that   A and       A    T    A have the same null space.We are given that   b is a non-zero vector such that       A    T    b  =  0. Assume there exists a vector   x such that   A  x  =  b. Multiply on the left by       A    T   to see that   x  ∈  N  (      A    T    A  ). It follows that   x  ∈  N  (  A  ). This implies that   b  =  0, which is a contradiction.Thus, the equation   A  x  =  b has no solutions.

Consider a system of linear equations A x = b where A is an n ×<!

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